Full Edge-Friendly Index Sets of One Point Union of Cycles

Zhen-Bin Gao1, Wai Chee Shiu2, Sin-Min Lee3, Gee-Choon Lau4
1College of General Education, Guangdong University of Science and Technology, Dongguan, 523000, P.R. China
2Department of Mathematics, The Chinese University of Hong Kong, Shatin, Hong Kong, P.R. China.
31786, Plan Tree Drive, Upland, CA 91784, USA
4College of Computing, Informatics & Mathematics, Universiti Teknologi MARA (Segamat Campus), 85000 Malaysia

Abstract

Let G=(V,E) be a graph with vertex set V and edge set E. An edge labeling f:EZ2 induces a vertex labeling f+:VZ2 defined by f+(v)uvEf(uv)(mod2), for each vertex vV. For iZ2, let vf(i)=|{vV:f+(v)=i}| and ef(i)=|{eE:f(e)=i}|. An edge labeling f of a graph G is said to be edge-friendly if |ef(1)ef(0)|1. The set {vf(1)vf(0):f is an edge-friendly labeling of G} is called the full edge-friendly index set of G. In this paper, we shall determine the full edge-friendly index sets of one point union of cycles.

Keywords: One point union of cycles, Edge labeling vector, Vertex labeling vector, Edge-friendly labeling, Full edge-friendly index set

1. Introduction

In this paper, all graphs are simple and connected. We refer to [1] for terms and notation that are not defined in this paper.

Since graph labelings was first introduced by Rosa, various labeling concepts have been introduced [2]. For example, cordial labeling, edge-friendly labeling, harmonious labeling, felicitous labeling, odd harmonious labeling and even harmonious labeling, semi-magic labeling, etc. Most graph labeling methods can be traced to the method introduced by Rosa [3] in 1967, or Graham and Sloane [4] in 1980.

Let G=(V,E) be a graph with vertex set V (or V(G)) and edge set E (or E(G)). A labeling f:EZ2 induces a vertex labeling f+:VZ2 defined by f+(v)uvEf(uv)(mod2), for each vertex vV. Here Z2 is the field of order 2.

For iZ2, let vf(i)=|{vV: f+(v)=i}| and ef(i)=|{eE : f(e)=i}|. The number vf(1)vf(0) is denoted by If(G) and is called the edge-friendly index of f.

An edge labeled by i is called an i-edge (under f). A vertex labeled by i is called an i-vertex (under f). In this paper, an edge labeling (or a (0,1)-edge labeling) means an edge labeling whose codomain is Z2. An edge labeling f of a graph G is said to be edge-friendly if |ef(1)ef(0)|1.

Definition 1. The full edge-friendly index set of G, denoted by FEFI(G), is the set {If(G) :f is an edge-friendly labeling of G}.

The concept of full edge-friendly index set was introduced in [5]. Since vf(1)+vf(0)=|V|, (1)If(G)=2vf(1)|V|=|V|2vf(0). So the edge-friendly index of f of G is determined by the value of vf(1).

Definition 2. Let f be an edge labeling of a graph G of order p and size q. We fix the sequence of vertices V={v1,v2,,vp} and the sequence of edges E={e1,e2,,eq} of G. The vector f(E)=(f(e1),f(e2),,f(eq))Z2q is called an edge labeling vector of f, where Z2n denotes the Cartesian product of n copies of Z2. Similarly, f+(V)=(f+(v1),f+(v2),,f+(vp))Z2p is called a vertex labeling vector of f.

For any (row) vector XZ2n, XT denotes the transpose of X.

Let M be the incidence matrix of a graph G according to fixed sequences of vertices and edges. Let f be an edge-labeling of G=(V,E). It is easy to see that f+(V)T=Mf(E)T, which is a column vector of length p over Z2.

For a vector XZ2p, the number of 1’s in X is denoted by wt(X), the Hamming weight of X. Hence, if f is an edge-friendly labeling of a graph G, then wt(f+(V)) is vf(1). For convenience, we will identify f with f(E)T. Hence vf(1)=wt(f+(V))=wt((Mf)T).

Let Hi be a graph and viV(Hi) be fixed, 1it. A one point union of Hi, 1it, is the graph obtained from the disjoint union of Hi by merging all vi into a single vertex which is called the merged vertex or core vertex. So there will be many different one point unions of Hi’s.

If Hi=Cni is a cycle of order ni3, then all one point unions of Hi, 1it, are isomorphic. So we denote the one point union of Cni, 1it, by k=1tCnk for t2. In this paper, we shall study the full edge-friendly index set of the graph k=1tCnk.

For 1kt, let Cnk=vk,1vk,2vk,nk1vk,nkvk,1. Let ek,j=vk,jvk,j+1 for 1jnk, where vk,nk+1=vk,1. Let vk,1V(Cnk) be the chosen vertex to be merged. We denote the core vertex of the graph Gt=k=1tCnk by c (or v0,0). Note that Gt is of order (k=1tnk)t+1 and of size k=1tnk. Following is the one point union of C3, C4 and C6, i.e., C3C4C6.

In the rest of the paper, we will use the notation defined above.

2. Some Extrema of Edge-Friendly Indices

In this section, necessary and sufficient conditions that give the extrema values of vf(0) or vf(1) are obtained. Following is Lemma 2.1 of [5,6].

Lemma 1. Let f be any edge labeling of a graph G=(V,E), then vf(1) must be even.

Suppose there are s odd numbers among n1,,nt, where 0st. Without loss of generality, we may assume that n1,,ns are odd and the others are even. Note that, s=0 means that there is no odd number; s=t means that there is no even number. Now (2)vf(1)+vf(0)=|V(Gt)|=k=1snk+k=s+1tnkt+1st+1(mod2). It is obvious that |V(Gt)|vf(0)0 for any edge labeling f of Gt. By Lemma 1 and the equation above, we have (3)vf(0)st+1(mod2). Hence vf(0)1 if ts is even for any edge labeling f of Gt. A natural question is that whether there is an edge-friendly labeling of Gt attaining the lower bound and whether there is an edge-friendly labeling of Gt attaining the upper bound.

Lemma 2. Suppose there are s odd numbers among n1,,nt. There is an edge-friendly labeling f of Gt such that

  1. vf(0)=1 if and only if ts is even;

  2. vf(0)=0 if and only if ts is odd.

Proof. The necessity of (1) and (2) come from Eq. (3).

Note that Gt is Eulerian. Let R be an Euler tour of Gt. We label the edge of Gt by 0 and 1 alternatively along R. Clearly the induced label of each vertex except the core is 1 and the label of the core c is (ts) mod2. Hence we have the sufficiency of (1) and (2). 0◻

Obtaining the maximum value of vf(0) is equivalent to obtaining the minimum value of vf(1)◻

Remark 1. If there are r numbers among n1,,nt such that the sum of these numbers is |E(Gt)|/2, then the sum of the remaining numbers is |E(Gt)|/2. So, the statement ‘there are r numbers among n1,,nt such that the sum of these numbers is |E(Gt)|/2’ is equivalent to the statement ‘there are r numbers among n1,,nt such that the sum of these numbers is |E(Gt)|/2’.

Lemma 3. There is an edge-friendly labeling f of Gt such that vf(1)=0 if and only if there are r numbers among n1,,nt such that the sum of these numbers is |E(Gt)|/2.

Necessity. Consider the subgraph Ei induced by all i-edges of Gt under f, i=0,1. Since there is no 1-vertex, each cycle of Gt is entirely edge-labelled by 1’s or entirely edge-labelled by 0’s. Hence E1 is a one point union of some subcycles of Gt. Hence E0 is also a one point union of some subcycles of Gt, namely E0 is the one point union of r subcycles. Since f is edge-friendly, |E(E0)|=|E(Gt)|/2 or |E(Gt)|/2. Thus we have the necessary condition by Remark 1.

[Sufficiency] Suppose there are r numbers among n1,,nt such that the sum of such numbers is |E(Gt)|/2. We label the edges of the corresponding cycles by 0 and label the other edges of Gt by 1. Thus this labeling is an edge-friendly labeling and the labels of all vertices are 0’s. ◻

3. Full Edge-Friendly Index Sets

The main results are given in this section. For a given one point union of cycles Gt=i=1tCni, we fix the sequences of vertices and edges with respect to the lexicographic order. Thus the incident matrix of Gt is M=(Z1Z2Zt1ZtB1OOOB2OOOBt1OOOOBt), where Bk=(110000011000001000000110000011)(nk1)×nk , Zk=(1001)1×nk, and each O is a zero matrix of certain size, 1kt.

We define some notation and recall some known results first. For positive l, let 1l be the row vector of length l whose entries are 1’s, and 0l be the row vector of length l whose entries are 0’s. Let α2l=(1,0,1,0,,1,0)Z22l and β2l=(0,1,0,1,0,1)Z22l. For convenience, we let 10, 00, α0 and β0 be the empty rows.

Let p=|V(Gt)|=(k=1tnk)t+1. From (2), we have pst+1(mod2), where s is defined in Section 2. So ts is odd if and only if p is even. From now on, we shall use q=k=1tnk=|E(Gt)|. For each edge labeling f=(Y1,Y2,,Yt)T, where YkZ2nk, 1kt, we have Mf=(k=1tZkYkT,B1Y1T,,BtYtT)T. From (1) we have FEFI(Gt)={2vf(1)p:f is an edge-friendlylabeling of Gt}{4jp:0jp/2} For each possible value 2j of the number of 1-vertices, we want to find an edge-friendly labeling f such that vf(1)=2j.

We deal with some special cases first.

Theorem 1. Suppose all n1,,nt are even.

  1. Suppose there are r numbers among n1,,nt such that the sum of these numbers is q/2, then

    1. FEFI(Gt)={4jp:0jp/2} if t is odd;

    2. FEFI(Gt)={4jp:0j(p1)/2} if t is even.

  2. Suppose the sum of any combination of integers n1,,nt is not equal to q/2.

    1. FEFI(Gt)={4jp:1jp/2} if t is odd;

    2. FEFI(Gt)={4jp:1j(p1)/2} if t is even.

Proof. For each i and l, 1it and 1lni/2, let Yi,l=(1l,0l,αni2l)Z2ni. Thus, BiYi,lT=(0l1,1,0l1,1ni2l)T and ZiYi,lT=1. Hence (4)wt(BiYi,lT)=ni2l+1. For 1kt, let f=(Y1,n1/2,,Yk1,nk1/2,Yk,l,Yk+1,1,,Yt,1)T. Thus,
Mf=(tmod2,B1Y1,n1/2T,,Bk1TYk1,nk1/2,BkTYk,l,Bk+1TYk+1,1,,BtTYt,1)T. By (4) (5)vf(1)=wt(Mf)=(tmod2)+i=1k11+(nk2l+1)+i=k+1t(ni1)

  1. Without loss of generality, we assume k=1rnk=q/2. Thus k=r+1tnk=q/2. Since either r or tr is at most t/2. We may assume that rt/2 and trt/2.

    • (1a) Suppose t is odd.

      Step 1: For a fixed k, 1kt, let l be an integer such that 1lnk/2.
      Let f be defined above, then (5) becomes
      vf(1)=wt(Mf)=1+i=k+1tni+2k+nk2lt=pi=1k1ni2l+2k. For a fixed k, when l runs through from 1 to nk/2, the range of vf(1) is an increasing arithmetic sequence with common difference 2 from pi=1kni+2k to pi=1k1ni2+2k. Thus when k runs through from 1 to t, the range of vf(1) is an increasing arithmetic sequence with common difference 2 from pi=1tni+2t=t+1 to p.

      Step 2: Let R be the Euler tour starts from the core c and travels the cycles Cn1, Cn2, , Cnt in order. Denote R=u1u2u3uq2uq2+1uq2+2uqu1, where u1=c. In this case uq2+1=c. There may have some ui’s equal to c. Let g=(1q2,0q2)T, then vg(1)=0.

      Step 3: Now we swap the labels of uiui+1 and uq2+2i1uq2+2i, 1iq2, where the indices are taken in modulo q. For each case, the number of 1-vertices increases by 2 if i=1 or c is not incident with neither uiui+1 nor uq2+2i1uq2+2i; and may not change if c is incident with either uiui+1 or uq2+2i1uq2+2i for i1. The last case can occur at most 2r1 times. So the number of 1-vertices increases by 2 at least q22r+12t2t/2+1t+1 times. Thus, the number of 1-vertices runs through each even integers from 0 to t+1 under the above swapping.

      So we obtain all the possible values of the number of 1-vertices.

    • (1b) Suppose t is even.

      Perform Step 1 as Case (1a). Now (5) becomes vf(1)=wt(Mf)=0+i=k+1tni+2k+nk2lt=pi=1k1ni2l+2k1. Similar to Case (1a), when k and l run through all possible values, vf(1) runs through p1, p3, …, t.

      Perform Steps 2 and 3 as Case (1a). We obtain that the number of 1-vertices runs through each even integers from 0 to t.

      So we obtain all the possible values of the number of 1-vertices.

  • By Lemma 3, 1vf(1)p for any edge-friendly labeling f. Without loss of generality, we assume that n1n2nt and k=1rnk<q/2<k=r+1tnk. Here rt/2.

    • (2a) Suppose t is odd. Do the same Step 1 as Case (1a) to obtain the edge labeling f. Thus, we obtain that vf(1) runs through p, p2, …, pi=1tni+2t=t+1.

      Let R be an Euler tour as in Case (1a). In this case, uq2+1c. Constructing the edge labeling g as in Case (1a), we get that vg(1)=2, since g+(u1)=g+(c)=1.

      Do the same Step 3 as in Case (1a). Here we obtain that the number of 1-vertices runs through each even integers from 2 to t+1.

      So we obtain all the possible values of the number of 1-vertices.

    • (2b) Suppose t is even. By a similar procedure and argument, we obtain all possible values of the number of 1-vertices.

  • Hence this completes the proof. ◻

    Theorem 2. Suppose all n1,,nt are odd.

    1. Suppose there are r numbers among n1,,nt such that the sum of these numbers is q/2, then
      FEFI(Gt)={4jp:0j(p1)/2}.

    2. Suppose the sum of any combination of integers n1,,nt is not equal to q/2, then
      FEFI(Gt)={4jp:1j(p1)/2}.

    Proof. Recall that α2l=(1,0,,1,0)Z22l and β2l=(0,1,,0,1)Z22l; 10, 00, α0 and β0 are the empty rows (see the beginning of this section).

    For a fixed i, 1it/2, let l be an integer such that 1l(ni1)/2. Let Yi,l=(1l,0l1,βni+12l)Z2ni, then BiYi,lT=(0l1,1,0l1,1ni2l)T and ZiYi,lT=0. Hence wt(BiYi,lT)=ni2l+1.

    For a fixed i, t/2+1it, let l be an integer such that 1l(ni1)/2. Let Yi,l=(0,1l,0l,αni12l)Z2ni, then BiYi,lT=(1,0l1,1,0l1,1ni12l)T and ZiYi,lT=0. Hence wt(BiYi,lT)=ni2l+1.

    Step 1: Let f=(Y1,(n11)/2,,Yk1,(nk11)/2,Yk,l,Yk+1,1,,Yt,1)T, where 1kt and 1l(nk1)/2. Clearly f is edge-friendly. vf(1)=wt(Mf)=i=1k12+(nk2l+1)+i=k+1t(ni1)=pi=1k1ni2l+3k2. One may check that when k and l run through all possible values, vf(1) runs through p1, p3, …, pi=1tni+3t1=2t.

    Step 2: For 1it/2, we define Yi,(ni+1)/2=(1(ni+1)/2,0(ni1)/2), then BiYi,(ni+1)/2T=(0(ni1)/2,1,0(ni3)/2)T and ZiYi,(ni+1)/2T=1. Hence wt(BiYi,(ni+1)/2T)=1.

    For t/2+1it, we define Yi,(ni+1)/2=(1(ni1)/2,0(ni+1)/2). Here BiYi,(ni+1)/2T=(0(ni3)/2,1,0(ni1)/2)T and ZiYi,(ni+1)/2T=1. Hence wt(BiYi,(ni+1)/2T)=1.

    For each k, 1kt, let fk=(Y1,(n1+1)/2,,Yk,(nk+1)/2,Yk+1,(nk+11)/2,,Yt,(nt1)/2)T. Note that, ft=(Y1,(n1+1)/2,,Yt,(nt+1)/2)T. Clearly fk is edge-friendly. Suppose k is even. We have vfk(1)=wt(Mfk)=({i=1k1mod2}+k)+2(tk)=(0+k)+2(tk)=2tk.

    One may check that when k runs through all even numbers from 2 to t, vfk(1) runs through all even numbers from 2t2 down to t.

    Now we shall perform some steps similar to Steps 2 and 3 in the proof of Theorem 1 to obtain the remaining possible values of the number of 1-vertices.

    1. Without loss of generality, we assume that k=1rnk=q/2. Here rt/2.

      Let R=u1u2u3uq/2uq/2+1uq/2+2uqu1 be the Euler tour starts from the core u1=c and travels the cycles Cn1, Cn2, , Cnt in order. By convention uq+1=u1.

      Step 3: Define g=(1q/2,0q/2)T. Now vg(1)=0.

      Step 4:

      1. When q=4m+1, where m2. Note that u2m+1=c, 4m+1=i=1tni3t and t is odd. Swap the labels of uiui+1 and u2m+2i1u2m+2i, 1im+1. The number of 1-vertices increases by 2 at least (m+1)(r1) times, since the first swapping increases vg(1) by 2.

        Next, swap the labels of um+1+ium+2+i and u2i1u2i, 1im1. So the number of 1-vertices increases by 2 at least m1r times.

        By the same argument as Step 3 in the proof of Theorem 1, the number of 1-vertices increases by 2 at least 2m+12r(3t1)/2+1t=(t+1)/2 times totally. Thus, the number of 1-vertices runs through each even integer from 0 to t+1 under the above swapping.

      2. When q=4m+3, where m2 (in this case q cannot be 7). Note that u2m+2=c, 4m+3=i=1tni3t and t is odd.

        Swap the labels of uiui+1 and u2m+2iu2m+2i+1, 1im+1. Next, swap the labels of um+1+ium+2+i and u2i1u2i, 1im.

        Totally, the number of 1-vertices increases by 2 at least (2m+1)2r+1(3t3)/2+2t=(t+1)/2 times. Thus, the number of 1-vertices runs through each even integer from 0 to t+1 under the above swapping.

      3. When q=4m, where m3 (in this case q cannot be 8). Note that u2m=c, 4m=i=1tni3t and t is even.

        Swap the labels of uiui+1 and u2m+2i1u2m+2i, 1im. Next, swap the labels of um+ium+i+1 and u2i1u2i, 1im1.

        Totally, the number of 1-vertices increases by 2 at least (2m1)2r+13t/2t=t/2 times. Thus, the number of 1-vertices runs through each even integer from 0 to t under the above swapping.

      4. When q=4m+2, where m1. Note that u2m+1=c, 4m+2=i=1tni3t and t is even.

        Swap the labels of uiui+1 and u2m+2iu2m+2i+1, 1im+1. Next, swap the labels of um+1+ium+i+2 and u2i1u2i, 1im.

        Totally, the number of 1-vertices increases by 2 at least (2m+1)2r+13t/2+1t=t/2+1 times. Thus, the number of 1-vertices runs through each even integer from 0 to t+2 under the above swapping.

    2. Without loss of generality, we assume that n1n2nt and k=1rnk<q/2<k=r+1tnk. Here rt/2. We do the same procedure as Case 1. The only difference is vg(1)=2. Hence the number of 1-vertices at least runs through each even integer from 2 to t under the above swapping.

    This completes the proof. ◻

    Example 1. Consider the graph C5C3C3C3C3. Here t=5, p=13, q=17, m=4 and r=2. Let R=u1u2u17u1 be an Euler tour of the graph, where u1=u6=u9=u12=u15=c. For 1i17, let ei=uiui+1, where u18=u1.

    The procedure of Steps 2 listed below: e1e2e3e4e5e6e7e8e9e10e11e12e13e14e15e16e17v(1)f11110010110101001010f2111001101010100108f3111001101100100108f4111001101101000106f5111001101101001006

    The procedure of swapping (Steps 3 and 4) listed below start from g=(18,09).
    e1e2e3e4e5e6e7e8e9e10e11e12e13e14e15e16e17v(1)g11111111000000000001111111100000000200111111101000000400011111101010000600001111101010100600000111101010101610000011101010101810100001101010101101010100010101010110

    Example 2. Consider the graph C3C3. Here t=2, p=5, q=6, m=1 and r=1. Let R=u1u2u6u1 be an Euler tour of the graph, where u1=u4=c. For 1i6, let ei=uiui+1, where u7=u1.

    The procedure of Steps 3 and 4 listed below start from g=(13,03). e1e2e3e4e5e6v(1)g1110000011100200110141001014

    Lemma 4. Let G and H be two graphs with only one common vertex u. Suppose fG and fH be two edge-friendly labelings of G and H, respectively. If (fG+(u),fH+(u))(1,1), then there is an edge-friendly labeling h of GH such that vh(1)=vfG(1)+vfH(1). If (fG+(u),fH+(u))=(1,1), then there is an edge-friendly labeling h of GH such that vh(1)=vfG(1)+vfH(1)2.

    Proof. Let h be the combined labeling of fG and fH. We obtain the lemma easily. ◻

    Theorem 3. Let p and q be the order and the size of Gt, respectively.

    1. Suppose there are r numbers among n1,,nt such that the sum of these numbers is q/2, then FEFI(Gt)={4jp:0jp/2}.

    2. Suppose there are no r numbers among n1,,nt such that the sum of these numbers are q/2, then FEFI(Gt)={4jp:1jp/2}.

    Proof. Without loss of generality, we assume that n1,,ns are even and ns+1,,nt are odd, where 0st. When s=0 or s=t, we get the results using Theorem 1 and Theorem 2.

    By Lemma 3, it suffices to find an edge-friendly labeling f of Gt such that vf(1) runs through all even numbers in [2,p].

    Let G=i=1sCni and H=i=s+1tCni. Let pG and pH be orders of G and H, respectively. Note that pH is odd and p=pG+pH1.

    By Theorem 1 there is an edge-friendly labeling fG of G such that vfG(1) at least runs through all even numbers of [2,pG], and by Theorem 2 there is an edge-friendly labeling fH of H such that vfH(1) at least runs through all even numbers of [2,pH1].

    By Lemma 4, in the worst case, there is an edge-friendly labeling h of GH such that vh(1) runs through all even numbers of [4,pG+pH12]=[4,p2].

    Now we only need to find an edge-friendly labeling g of GH such that vg(1) is p when p is even, or p1 when p is odd, or else, vg(1) is 2 (see 1⃝ and 2⃝ below).

    Let R=u1u2u3uq/2uq/2+1uq/2+2uqu1 be the Euler tour of Gt starts from the core u1=c. Note that q=p+t1 and deg(c)=2t.

    1. When q is even, pt1(mod2). Define g=(1,0,1,0,,1,0)TZ2q, then g+(c)=tmod2. Hence vg(1)={pif t isodd,p1if t is even,={pif p is even,p1if p is odd.

      When q is odd, pt(mod2). Define g=(1,0,1,0,,1,0,1)TZ2q, then g+(c)=t+1mod2. Hence vg(1)={p1if t isodd,pif t is even,={p1if p is odd,pif p is even.

    2. Suppose uq/2+1c. Define g=(1q/2,0q/2)T. Thus only g+(c)=g+(uq/2+1)=1. Hence vg(1)=2.

      Suppose uq/2+1=c, then u2c and uq/2+2c. Define g=(0,1q/2,0q/21)T. Thus only g+(u2)=g+(uq/2+2)=1. Hence vg(1)=2.

    This completes the proof. ◻

    We denote the graph Gt by Cn(t) when n1==nt=n. When n=3, C3(t) is called the Dutch t-windmill graph. By Theorem 3, we have the following corollaries.

    Corollary 1. Suppose n3 and t2. Now, the order of Cn(t) is p=ntt+1.

    1. Suppose n is odd, then FEFI(Cn(t))={{4jp:1j(p1)/2} if t is odd;{4jp:0j(p1)/2} if t is even.

    2. Suppose n is even, then FEFI(Cn(t))={{4jp:1jp/2} if t is odd;{4jp:0j(p1)/2} if t is even.

    Corollary 2. For t2, FEFI(C3(t))={{4j2t1:1jt} if t is odd;{4j2t1:0jt} if t is even.

    Funding

    The first author was partially supported by National Natural Science Foundation of China (No: 12371344).

    Financial Interests

    The authors have no relevant financial or non-financial interests to disclose.

    Author Contributions

    All authors contributed to the study conception and design. Material preparation and analysis were performed by all the authors. The first draft of the manuscript was written by Zhen-Bin Gao and all authors commented on previous versions of the manuscript. All authors read and approved the final manuscript.

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